What is the integral from 1 to 3 of (x^2+2x-3) ?

The integral from 1 to 3 of (x^2+2x-3) is 32/3

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integral from 1 to 3 of (x^2+2x-3)

\int_{1}^{3} (x^{2} + 2 x - 3) d x

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

\int_{1}^{3} x^{2} + 2 x - 3 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} x^{2} d x + \int_{1}^{3} 2 x d x - \int_{1}^{3} 3 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 2 x d x = 8

\int_{1}^{3} 3 d x = 6

= \frac{26}{3} + 8 - 6

Simplify

\frac{26}{3} + 8 - 6 : \frac{32}{3}

= \frac{32}{3}

\int x^{3} arcsin (x) d x

2.5 x^{2} + 0.36 + 4 (1.1 - x)

\int \frac{x + 4}{x ^{2} + 9 x - 10} d x

f (x) = - \frac{2}{x ^{3}}

n = 0 \sum \infty (\frac{n}{n + 2})^{2}

What is the integral from 1 to 3 of (x^2+2x-3) ?
The integral from 1 to 3 of (x^2+2x-3) is 32/3